The third factor on which my ionization enthalpy depends is the penetration effect. Now penetration over here does not have the literal English meaning of penetrating through something or passing through something. Penetration just means ‘the proximity of an electron in an orbital to the nucleus’ or we can also say that how effectively can an electron get close to the nucleus. Now the penetration of the nucleus by an electron is measured by the relative density near the nucleus of an atom for each shell and subshell of an e. now if we recall the radial probability distribution functions that we had started in quantum numbers, the RDF plots for the second shell would look something like this. Right? Now over here if you see that the electron density of the 2s electrons is closer to the nucleus as compared to the 2p. So that means the electron density of a 2s electron is more towards the nucleus as compared to 2p. So that means the 2s electrons or the 2s orbital is penetrating towards the nucleus more as compare to the 2p. Similarly, if I take the plot for the third shell, you will clearly see that the electron density of the 3s is closer to the nucleus as compared to the 3p and 3d. So in general can I conclude that the penetration power of the orbitals would be in the order that S would have the highest penetration followed by P, D and F? Which is very clearly verified by the radial distribution functions graph that we have. Now using this what do we do? Now if I arrange them in the increasing order, so how will the penetration effect look like? So the penetration power of the orbitals would have 1s would be the closest to the nucleus followed by 2s, 2p, 3s, 3p, 4s, 3d. Now, over here there is a discrepancy. Because we know that in a shell it should have been s p d. But why is the s of fourth shell more penetrating that is again depending on the mathematical calculations of the quantum number that we got. We easily saw that the 4s was much more penetrating towards the nucleus as compared to 3d. So this was the order. Now if I reverse this it will actually give me how strongly is one orbital bound to the nucleus. So in total we can find out which will have more energy. Now if 1s is bound strongest to the nucleus that means that it will have the least amount of energy and the electron present in 1s would be the most stable. But as we go on towards the left side, an electron in 3p would be far away from the nucleus. That means the energy of an electron in 3p would be very high and the electron would be unstable. So this would be my penetration effect sequence for the orbitals and just the reverse of it would give me the energies associated with each of the orbitals. Now if you recall one more thing the bolsberry rule that we studied in quantum number that was the n+ l rule, it exactly satisfies this sequence. So this is the penetration effect. So that means an electron, if I am talking about an electron in the 3rd shell, if it is present in 3s orbital it would be bound strongly to the nucleus as compared to the electron in 3p or 3d of the same shell. So it would be very difficult for us to remove an electron from 3s orbital as compared to the 3p. So that is penetration effect.
So the last factor on which my ionization enthalpy depends is the shielding effect. Now as soon as you hear the word shielding there would be some image that gets created in your mind. Let’s say if I have two boxers who are fighting and the fight goes out of control then what happens – the referee has to walk in and separate the 2 boxers. So what exactly is the referee doing? Referee is acting as a shield to move away the boxers. Now is that what happens in the atom. No. In the atom, the shielding has a totally different meaning. Where, let’s try to understand this using Shah Rukh Khan. You must all be familiar who Shah Rukh Khan is. He is a very famous personality. Now what happens? Now if I have Shah Rukh Khan in the center and to protect Shah Rukh Khan there would be his security which would be surrounding him. Right? Now let’s say if the security allows 50 people to come towards Shah Rukh Khan and have a look at him, take his pictures and all, so what would those 50 people do? They would make a circle around Shah Rukh Khan but they cannot come close enough to Shah Rukh Khan because their security guards are already there who are trying to shield Shah Rukh Khan from those 50 people who’ve come in. Now let’s assume that the security guards allow another 50 people to come in. Now what happens? The first 50 people who are already surrounding Shah Rukh Khan, would they want the other 50 people to come as close as they are to Shah Rukh Khan? No. So that is what is happening that the security guard is shielding the first 50 people and that first 50 people are now shielding the second 50 people. So this is how the shielding is happening and the shielding over here is that they are not allowing the people to go close to Shah Rukh Khan and have a look at him. So this is the same thing which is happening in the case of atoms that the electrons which are present before the other electrons which are towards the outer side, the inner electrons do not allow the appropriate nuclear charge to go towards the outermost electrons. Now, for this, let’s say I have the nucleus and around the nucleus there would be a field of nuclear charge, or as the physics people would say the electrostatic potential field. Now what happens in that electrostatic potential field that as soon as an electron comes in because of the electrostatic forces of attraction with the nucleus they would be attracted? Right? Now closer the electron is to the nucleus, more would be the attractive force from the nucleus on the electron and the energy of that electron would go down, meaning the electron becomes more stable. So what happens? All the electrons would want to be as close to the nucleus. Now think about a neutral atom. For the neutral atom the electron to proton ration should be 1 is to 1. That means each proton is holding onto each electron. But still – but still some of the electrons in an atom are more strongly held to the nucleus as compared to the others. That is because of the distance. The closer the electron to the nucleus, it takes up the nuclear charge more effectively but when other electrons try to come in, the other shell, these electrons which are closer to the nucleus act as a shield and do not allow the appropriate nuclear charge to be felt by the electron. Hence, outer electrons experience the effective nuclear charge and not the actual nuclear charge and this effective nuclear charge is a result of the shielding which is happening because of the electrons present before the outer electrons. So this effective nuclear charge is denoted by a term called as Zeffective, and, Zeffective is given as Zeffective is equal to Z – S, or, sometimes it is also called as sigma. Now this Zeffective is the effective nuclear charge. Z is the total number of protons or the total nuclear charge. And S, or sigma, is the screening constant. Now this screening constant, the value that you will be finding out for the screening constant is found out using a rule called as the Slater’s Rule. Now this Slater’s Rule is something that you would understand when you are doing your Master’s in Chemistry and all so this Slater’s Rule is not at all recommended to be studied over here. But, now as chemistry is advancing, what has happened is that the Slater’s Rule is also a semi empirical formula, i.e., it’s not a foolproof method of finding the screening constant. So what happened was, there are two scientists, Clementine and Rymondie who came into the picture and gave us the values of the effective nuclear charge on different orbitals as this. Right? Now if you see the values – the values when you are going from left to right, the values are increasing for the effective nuclear charge on 1s, 2s and similarly 2p. Right? Now let’s keep this table aside for some time and let us understand what exactly is happening. Now if I have a 1 electron system, so what happens for a 1 electron system – I have 1s subshell, 2s, 2p, 3s, 3p and 3d. Now for each shell the subshells are degenerate, i.e., they have the same energy. Right? So now what happens is if there is a 1 electron system then this 1s subshell is the one which is shielding the nuclear charge for the second shell. So that means the attractive force or the nuclear charge experienced by 2s and 2p should be equal because 1s is shielding the second shell completely. And similarly, this 2s and 2p is shielding the 3s and 3p or 3d equally. So the values experienced by the subshells, the orbitals, the electrons in 2s, 2p should be same and similarly for 3s, 3p and 3d it should be same. But now the problem arises that if you take a look at what Clementie and Rymondie observed, they have seen that the value of 1s is increasing. The value of 2s is also increasing. Now the value of 2p – it should have been same, but it is different for 2p, because now, that means 1s is shielding the 2s and 2s is shielding the 2p. Now how do we justify that? Now this actually is for a multi-electron system. Now in a multi-electron system what happens – there is 1s, 2s, 2p, 3s, 3p, 3d. Let’s say we have these subshells. Now if you recall the penetration effect that we just did, what happens? 1s is closer to the nucleus, then in the second shell 2s penetrates more as compared to 2p, so that means 2s is closer as compared to 2p and in the third shell 3s is closer as compared to 3p and 3d. So what it effectively means that 1s is shielding 2s and the 2s is now shielding the 2p and similarly 1s, 2s, 2p are shielding the 3s and for the same shell 3s is shielding 3p and 3d. So that is why these values that you see have a different value and they decrease as you move to different subshells. So 1s, for say, carbon has this value and 2s has a lower value. Now if we move to nitrogen, the 1s is having a value, 2s is having a lower value and the 2p is having an even lower value. But if you compare the 1s of carbon and nitrogen, the 1s has an effective nuclear charge more for nitrogen as compared to carbon. Now why is that happening? Because the total nuclear charge, or the number of protons present for carbon are 6 and for nitrogen it’s 7. So that means the total nuclear charge is also increasing and hence if you apply the Zeffective is equal to Z – S value, the value, i.e., the Zeffective value for each subshell would increase and hence the subshell where the effective nuclear charge is more, so I would need more amount of energy to break the interaction, or break the interactive force that the nucleus has with it and the ionization enthalpy would increase. So this is what shielding effect is.
And the last factor which effects ionization enthalpy is the electronic configuration. Now over her you should recall we have done something as half-filled and fully filled orbitals or the subshells where we said that half-filled and fully filled subshells have extra stability. So if I try to remove an electron from a half filled or fully filled subshell what I am trying to do? I am trying to decrease the stability so would the atom want to decrease the stability. It won’t. Right? So that is why it will demand more energy from our side to remove the electron and hence the ionization enthalpy of such atoms would be higher as compared to what value they should have had. So these are the 5 factors on which the ionization enthalpy depends.
Now after understanding all the factors let us talk about the periodic trends. Now the periodic trends, if i move from left to right in a period, what happens? The atomic radius decreases, the size that is decreasing, so if the size is decreasing, the amount of attractive force from the nucleus to the outermost electron is increasing. Right? So the ionization enthalpy should also increase, because I would need more amount of energy to break the higher attractive forces. So, if I am moving from left to right in a period the ionization enthalpy should increase. And, it is the opposite when I am going down the group. If I go down the group, the ionization enthalpy would be decreasing because the size is increasing and as a consequence the amount of attractive force is decreasing.
So, now let’s take the period which contains lithium till neon and let us see are they actually following the periodic trend that I told you or are they behaving weirdly. Right? So let’s start from lithium. So in lithium, the electronic configuration that you have is 1s2, 2s1 and with beryllium it is 1s2, 2s2. Now, since the number of electrons has increased, so the size would be decreasing and if the size has decreased the attractive force has increased and more amount of energy would be required to remove an electron from beryllium as compared to lithium. So the ionization enthalpy is increasing as you are going from left to right. But now when we come to boron – what happens? Boron has an electronic configuration of 1s2, 2s2, 2p1. Right? So the atomic radius of boron is smaller as compared to beryllium so the ionization enthalpy of boron should be higher. But no. That’s not the case. The ionization enthalpy of beryllium is higher as compared to boron because whenever we are talking about the ionization enthalpy all the 5 factors have to be taken into account. It’s not only on 1 factor where we can answer the question. So now on the basis of atomic radius, boron should have a higher ionization enthalpy, but on the basis of electronic configuration beryllium should have the higher ionization enthalpy because it has a fully filled s subshell. The second reason for beryllium having a higher ionization enthalpy is the penetration effect, because the 2s is the most closely bound in the second shell to the nucleus but in boron you have 2s and 2p also. The 2p electron is not so closely bound to the nucleus as compared to 2s. So removing an electron or removing the 2p electron from boron becomes easier as compared to removing the s electron from beryllium. So these are the two reasons why beryllium has a higher ionization enthalpy as compared to boron. Now, if we move ahead with boron, carbon and nitrogen the ionization enthalpy should be increasing because the atomic radius is decreasing. Right? And there is no anomaly based on the electronic configuration or penetration that is happening over here. Now if we compare nitrogen and oxygen, what happens? There is another anomaly that happens over here. Because if you take a look at the electronic configuration of nitrogen it has 2p3 which is again a half filled subshell. But with oxygen it is not the case. So again, using the same funda as we used for beryllium and boron over here also, nitrogen will have a higher stability and some extra amount of energy would be required to remove the electron from nitrogen. So the ionization enthalpy of nitrogen is more as compared to oxygen. Now, from oxygen if we go ahead to fluorine and neon, again there is no anomaly as such because the size keeps on decreasing so the amount of ionization enthalpy would be increasing. Now in the case of neon, neon is a noble gas which has a fully filled orbit. Now no matter how high energy you give, can you ionize neon? It won’t be possible because you are going against the laws of chemistry where you are destabilizing and atom which is again not possible. So these are the anomalies or the exceptions that happen in the second period. Now, similarly if you move to the third period, fourth period and fifth period you can take into account the anomalies that I discussed, the reasons that I discussed and you can find out the answers for that.
Now since we have got a fair understanding of what ionization enthalpy is a better understanding would be when you start solving questions. The questions based on ionization enthalpy are all there in the question/answer module. Over here I will be discussing questions which require the actual understanding. Now, the first question that I am going to discuss is this. So now let’s take a look at how to approach such a question. If you see the value, the ionization enthalpy value 1 is given as 403 for A, 549 for B, and 1142 for C. So which is the lowest ionization enthalpy value over here. It’s actually for A. Right? Now that means the first electron can be easily removed from A, then it is B and finally it is C. Now let’s take a look at the ionization enthalpy value 2. The first value for A is 2640, for B it is 1060 and for C it is 2080. So where do you see that a difference of more than 2100 is there because as a jugaad I can say that wherever you have a value of 2100 difference, of if there is a difference of 2100 between the two ionizations, that is the point where the noble gas configuration is reached. So if the first ionization enthalpy value for A is 403 and the second value is 2640 the difference is more than 2100, so that means it should be an element belonging to the first group, which is the alkali metal. Because if I take sodium, for e.g., I can remove 1 electron very easily, but after removing that electron it achieves a noble gas and it belongs to the first group. So for the first group elements, the first ionization enthalpy is very low and the second ionization enthalpy would be very high. So element A is belonging to alkali metal. Now if I talk about element B, the first value is 549 and the second value is 1060. That means removing the first electron is a little difficult from A and removing the second electron is also kind of easy. So this is the pattern, or this is the element where it should belong to the second group. Because in the second group you can ionize two electrons and both the values are somewhat identical and they don’t have a difference of more than 2100 which indicates that noble gas configuration has yet not been achieved. Now if we talk about C, if you check the value, the ionization enthalpy 1 is given to as 1142 and the second ionization enthalpy value is given as 2080. Now if you see the first ionization enthalpy value it is almost the double of the either A or B so that means, it is not very easy to remove an electron from C. That means, it indicates that it is a non-metal because non-metals are the ones which do no remove electrons, rather they want to accept electrons and metals are the ones which love to remove electrons. That is why there first ionization enthalpy values are low. So, just by looking at the value of ionization enthalpy 1 and comparing it to the other two values you can easily say that you element C is a non-metal. So this is how you do such a question.
Now let’s take an example. Let’s do the second question. The second question looks something like this. Now in this question they are asking you to find out noble gas, the reactive metal, the reactive non-metal and a metal which can form a binary halide. So that means a binary halide is the one which would only get two ionization enthalpies and that is something belonging to group 2. Right? Now first let us indicate the noble gas. Which one is the noble gas? Now whichever has the highest first ionization enthalpy value and the second ionization enthalpy value would be a noble gas because it is the highest amount of energy, which is required in noble gases. So if you take a look, the first ionization enthalpy value of element 1 is given as 2372 and for 2 it is given to as 5251. So both values are very high. So element 1 is always going to be a noble gas element. Now if we talk about the second, third and the fourth, out of the 3 which one has the lowest first ionization enthalpy value? The one which has the lowest first ionization enthalpy value would be the one which would be acting as a metal. Because metals, again, want to give away electrons. So out of the remaining 3, element B would be, or element 2 would be your metal. Right? Now let’s see 3 and 4. Now we are left with one of the elements would form a binary compound and the other one would form a non-metal. Now out of the two, whichever has a higher ionization enthalpy would be a non-metal. So we have the values of 900 and 1680. So which one do you think should be a non-metal? Yes, element 4 is a non-metal and we are left with element 3. Now in element 3 if you look carefully the ionization enthalpy values are given as 900 and 1760. So that means I am removing one electron easily and the second electron is also being removed kind of easily as compared to the other ones. So I can remove two electrons, i.e., I will have two positive charge and with a halide I can form a binary compound. So element 3 would be the element which belongs to group 2. So this is how you solve the questions taking into account the ionization enthalpy values.